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ent 5.95 years. b. Assume 80 Cheek My
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Here, we perform T-test for One Population Mean :

The provided sample mean is X 9.44 and the sample standard deviation is s 5.95, and the sample size is n-80 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested Ho: μ 10 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is α-0.01, and the critical value for a left- tailed test is te2.374 The rejection region for this left-tailed test is R-t:t<-2.374 (3) Test Statistics The t-statistic is computed as follows t Λ-140 9.44-10 s Vn 5.95/80 0.842(4) Decision about the null hypothesis Since it is observed that t =ー0.842 > tc =ー2.374, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p = 0.2012, and since p-0.2012 > 0.01, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is less than 10, at the 0.01 significance level.

Conclusion : Do not reject the null hypothesis. We do not have enough evidence to disprove the shareholder group's claim.

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